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922. Sort Array By Parity II

Description

Given an array of integers nums, half of the integers in nums are odd, and the other half are even.

Sort the array so that whenever nums[i] is odd, i is odd, and whenever nums[i] is even, i is even.

Return any answer array that satisfies this condition.

 

Example 1:

Input: nums = [4,2,5,7]
Output: [4,5,2,7]
Explanation: [4,7,2,5], [2,5,4,7], [2,7,4,5] would also have been accepted.

Example 2:

Input: nums = [2,3]
Output: [2,3]

 

Constraints:

• 2 <= nums.length <= 2 * 104
• nums.length is even.
• Half of the integers in nums are even.
• 0 <= nums[i] <= 1000

 

Follow Up: Could you solve it in-place?

Solutions

• Java
• C++
• Python
• Go
• Javascript
• TypeScript
• RenderScript

• class Solution {
      public int[] sortArrayByParityII(int[] nums) {
          for (int i = 0, j = 1; i < nums.length; i += 2) {
              if ((nums[i] & 1) == 1) {
                  while ((nums[j] & 1) == 1) {
                      j += 2;
                  }
                  int t = nums[i];
                  nums[i] = nums[j];
                  nums[j] = t;
              }
          }
          return nums;
      }
  }
  

• class Solution {
  public:
      vector<int> sortArrayByParityII(vector<int>& nums) {
          for (int i = 0, j = 1; i < nums.size(); i += 2) {
              if ((nums[i] & 1) == 1) {
                  while ((nums[j] & 1) == 1) {
                      j += 2;
                  }
                  swap(nums[i], nums[j]);
              }
          }
          return nums;
      }
  };
  

• class Solution:
      def sortArrayByParityII(self, nums: List[int]) -> List[int]:
          n, j = len(nums), 1
          for i in range(0, n, 2):
              if (nums[i] & 1) == 1:
                  while (nums[j] & 1) == 1:
                      j += 2
                  nums[i], nums[j] = nums[j], nums[i]
          return nums
  
  

• func sortArrayByParityII(nums []int) []int {
  	for i, j := 0, 1; i < len(nums); i += 2 {
  		if (nums[i] & 1) == 1 {
  			for (nums[j] & 1) == 1 {
  				j += 2
  			}
  			nums[i], nums[j] = nums[j], nums[i]
  		}
  	}
  	return nums
  }
  

• /**
   * @param {number[]} nums
   * @return {number[]}
   */
  var sortArrayByParityII = function (nums) {
      for (let i = 0, j = 1; i < nums.length; i += 2) {
          if ((nums[i] & 1) == 1) {
              while ((nums[j] & 1) == 1) {
                  j += 2;
              }
              [nums[i], nums[j]] = [nums[j], nums[i]];
          }
      }
      return nums;
  };
  
  

• function sortArrayByParityII(nums: number[]): number[] {
      for (let i = 0, j = 1; i < nums.length; i += 2) {
          if (nums[i] % 2) {
              while (nums[j] % 2) {
                  j += 2;
              }
              [nums[i], nums[j]] = [nums[j], nums[i]];
          }
      }
      return nums;
  }
  
  

• impl Solution {
      pub fn sort_array_by_parity_ii(mut nums: Vec<i32>) -> Vec<i32> {
          let n = nums.len();
          let mut j = 1;
          for i in (0..n).step_by(2) {
              if nums[i] % 2 != 0 {
                  while nums[j] % 2 != 0 {
                      j += 2;
                  }
                  nums.swap(i, j);
              }
          }
          nums
      }
  }
  
  

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